Differentiablity

1,2 and 3

Yes Aditiya...

\hspace{-16}\mathbf{f(x)=\mid x-1\mid ^3+\mid x-2\mid^5+\mid x-3\mid}\\\\\\ $So $\mathbf{f(x)=\begin{Bmatrix} \bold{-(x-1)^3-(x-2)^5-(x-3)}\;\;,& \bold{x\leq 1} \\\\ \bold{(x-1)^3-(x-2)^5-(x-3)}\;\;,& \bold{1 <x\leq 2} \\\\ \bold{(x-1)^3+(x-2)^5-(x-3)}\;\;,& \bold{2 <x\leq 3} \\\\ \bold{(x-1)^3+(x-2)^5+(x-3)}\;\;,& \bold{x> 3} \end{Bmatrix}}$\\\\\\\ $\mathbf{f^{1}(x)=\begin{Bmatrix} \bold{-3(x-1)^2-5(x-2)^4-1}\;\;,& \bold{x\leq 1} \\\\ \bold{3(x-1)^2-5(x-2)^4-1} \;\;,& \bold{1 <x\leq 2}\\\\ \bold{3(x-1)^2+5(x-2)^4-1} \;\;,& \bold{2 <x\leq 3}\\ \\ \bold{3(x-1)^2+5(x-2)^4+1}\;\;,& \bold{x> 3} \end{Bmatrix}}$\\\\\\ So Here $\mathbf{f^{1}(1^{-})=f^{1}(1^{+})}$\\\\ and $\mathbf{f^{1}(2^{-})=f^{1}(2^{+})}$\\\\ But $\mathbf{f^{1}(3^{-})\neq f^{1}(3^{+})}$